Question

If \( z^2 + z + 1 = 0 \) where \( z \) is a complex number, then the value of \( \left( z + \frac{1}{z} \right)^2 + \left( z^2 + \frac{1}{z^2} \right)^2 + \left( z^3 + \frac{1}{z^3} \right)^2 \) equals:

• A. \( 4 \)
• B. \( 5 \)
• C. \( 6 \)
• D. \( 7 \)
• E. \( 8 \)

Hint:

For any expression involving \( z^n + 1/z^n \) where \( z^2+z+1=0 \), the value repeats every 3 terms. If \( n \) is a multiple of 3, the sum is 2; otherwise, the sum is -1.

Answer 1

The Correct Option is C

Concept: The equation \( z^2 + z + 1 = 0 \) defines the non-real cube roots of unity (\( \omega \) and \( \omega^2 \)). These roots satisfy \( z^3 = 1 \) and \( z + \frac{1}{z} = -1 \). We can use these properties to evaluate each term in the sum.

Step 1: Evaluating each squared term.
From \( z^2 + z + 1 = 0 \), divide by \( z \): \( z + 1 + \frac{1}{z} = 0 \implies z + \frac{1}{z} = -1 \). \[ \text{Term 1:} \quad \left( z + \frac{1}{z} \right)^2 = (-1)^2 = 1 \] For Term 2, \( z^2 + \frac{1}{z^2} = (z + \frac{1}{z})^2 - 2 = (-1)^2 - 2 = -1 \). \[ \text{Term 2:} \quad \left( z^2 + \frac{1}{z^2} \right)^2 = (-1)^2 = 1 \] For Term 3, since \( z^3 = 1 \): \[ \text{Term 3:} \quad \left( 1 + \frac{1}{1} \right)^2 = (2)^2 = 4 \]

Step 2: Summing the results.
\[ \text{Total Value} = 1 + 1 + 4 = 6 \]